Select your answer What is $$ \mathrm{i}^{17} $$ (3 out of 6) $$ \sqrt{-1} $$ -1 1 It cannot be determined.

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The value of $$ \mathrm{i}^{17} $$ can be determined by recognizing the cyclical pattern of powers of $$ \mathrm{i} $$:

$$
\begin{align*}
\mathrm{i}^1 &= \mathrm{i} \
\mathrm{i}^2 &= -1 \
\mathrm{i}^3 &= -\mathrm{i} \
\mathrm{i}^4 &= 1 \
\mathrm{i}^5 &= \mathrm{i} \
&\vdots
\end{align*}
$$

This cycle repeats every 4 powers. To find $$ \mathrm{i}^{17} $$, divide 17 by 4:

$$
17 \div 4 = 4 \text{ with a remainder of } 1
$$

So,

$$
\mathrm{i}^{17} = \mathrm{i}^{4 \times 4 + 1} = (\mathrm{i}^4)^4 \times \mathrm{i} = 1^4 \times \mathrm{i} = \mathrm{i}
$$

Among the given options, $$ \sqrt{-1} $$ is equal to $$ \mathrm{i} $$.

**Answer:** $$ \sqrt{-1} $$
$$ \mathrm{i}^{17} = \sqrt{-1} $$

Select your answer What is \( \mathrm{i}^{17} \) (3 out of 6) \( \sqrt{-1} \) -1 1 It cannot be determined.

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